How do you evaluate tan^-1(tan(pi)) without a calculator?

Sep 28, 2016

${\tan}^{-} 1 \left(\tan \pi\right) = 0$.

Explanation:

First of all, recall that $\tan \pi = 0$, so, the reqd. value is ${\tan}^{-} 1 \left(0\right)$

To find this, we must know the following Defn. of ${\tan}^{-} 1$ fun. :

${\tan}^{-} 1 x = \theta , x \in \mathbb{R} \iff \tan \theta = x , \theta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$.

Now, knowing that, $\tan 0 = 0 , \mathmr{and} , 0 \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, we can conclude

from the Defn. that, ${\tan}^{-} 1 0 = 0 , i . e . , {\tan}^{-} 1 \left(\tan \pi\right) = 0$.

It will be interesting to note that, ${\tan}^{-} 1 \left(\tan \pi\right) \ne \pi$.